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Tích 1.2.3....2009 luôn chứa thừa số 10
=> Tất nhiên sẽ có chữ số tận cùng là 0
Vì \(\frac{1}{3}A=\frac{2}{5}B\Leftrightarrow A=\frac{2}{5}B\div\frac{1}{3}=\frac{2}{5}B\cdot3=\frac{6}{5}B\)
Vậy A = 6/5B
1/1x2 + 1/2x3+ 1/3x4+....+1/2009x2010
= 1/1-1/2 + 1/2-1/3+ 1/3-1/4+...+1/2009-1/2010
= 1/1-1/2010
= 2009/2010
\(\frac{1}{1\cdot2}+\cdot\cdot\cdot+\frac{1}{2009\cdot2010}\)
\(=1-\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}\)
\(=\frac{2009}{2010}\)
=1/2-1/3+1/3-1/4+...+1/a-1/(a-1)
=1/2-1/(a-1)
đề thiếu
\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
\(x=\dfrac{9}{12}-\dfrac{2}{12}\)
\(x=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{1}{2020}\)
\(x=\dfrac{1010}{2020}-\dfrac{1}{2020}\)
\(x=\dfrac{1009}{2020}\)
\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}-x\)
\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1\times2\times3\times4\times...\times2019}{2\times3\times4\times5\times...\times2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{2020}=\dfrac{1009}{2020}\)
A = (1 - \(\frac{1}{2}\)) x (1 - \(\frac{1}{3}\)) x (1 - \(\frac{1}{4}\)) x (1 - \(\frac{1}{5}\)) x ... x (1 - \(\frac{1}{2014}\)) x (1 - \(\frac{1}{2015}\))
A = \(\frac{1}{2}\)x \(\frac{2}{3}\) x \(\frac{3}{4}\) x \(\frac{4}{5}\) x ... x \(\frac{2013}{2014}\)x \(\frac{2014}{2015}\)
A = \(\frac{1x2x3x4x...x2013x2014}{2x3x4x5x...x2014x2015}\)
A = \(\frac{1}{2015}\)
Vậy A = \(\frac{1}{2015}\)
~~~
a) (456x35+65x456):19
=456x(65+35):19
=456x100:19
=45600:19
=2400
k mình nhé
Chúc bạn học giỏi
a=1/2.2+1/3.3+1/4.4+...+1/2009.2009+1/2010.2010(có 2009 số hạng)
a=1+1+1+...+1+1(2009 số 1)
a=1.2009=2009
Vậy a>1
https://scratch.mit.edu/projects/782275470